Divide the following complex numbers. $ \dfrac{-7-i}{-1-i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-1+i}$ $ \dfrac{-7-i}{-1-i} = \dfrac{-7-i}{-1-i} \cdot \dfrac{{-1+i}}{{-1+i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-7-i) \cdot (-1+i)} {(-1-i) \cdot (-1+i)} = \dfrac{(-7-i) \cdot (-1+i)} {(-1)^2 - (-1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-7-i) \cdot (-1+i)} {(-1)^2 - (-1i)^2} = $ $ \dfrac{(-7-i) \cdot (-1+i)} {1 + 1} = $ $ \dfrac{(-7-i) \cdot (-1+i)} {2} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-7-i}) \cdot ({-1+i})} {2} = $ $ \dfrac{{-7} \cdot {(-1)} + {-1} \cdot {(-1) i} + {-7} \cdot {1 i} + {-1} \cdot {1 i^2}} {2} $ Evaluate each product of two numbers. $ \dfrac{7 + 1i - 7i - 1 i^2} {2} $ Finally, simplify the fraction. $ \dfrac{7 + 1i - 7i + 1} {2} = \dfrac{8 - 6i} {2} = 4-3i $